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homework 2-solutions

homework 2-solutions - aleman(ka6539 homework 2...

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aleman (ka6539) – homework 2 – Cowley – (90225) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points What mass of O 2 is required to produce 14.5 g of CO 2 if the reaction has a 65.0% yield? CH 4 + 2 O 2 CO 2 + 2 H 2 O 1. 10.5 g 2. 5.27 g 3. 32.4 g correct 4. 16.2 g 5. 8.11 g 6. 13.7 g 7. 21.1 g Explanation: yield = 14 . 5 g % yield = 65% 14.5 g is the actual yield. We need to calculate the theoretical yield: 65 . 0 100 = 14 . 5 g CO 2 th . yield th . yield = 22 . 3 g CO 2 ? g O 2 = 22 . 3 g CO 2 × 1 mol CO 2 44 g CO 2 × 2 mol O 2 1 mol CO 2 × 32 g O 2 1 mol O 2 = 32 . 4 g O 2 002 10.0 points Which of the following would contain the largest number of moles? 1. 5.00 grams of calcium (Ca) 2. 5.00 grams of chlorine (Cl) 3. 5.00 grams of copper (Cu) 4. All the answers contain the same number of moles. 5. 5.00 grams of lithium (Li) correct Explanation: The one with the largest number of moles would need to have the smallest atomic weight. In this case, it would be lithium. 003 10.0 points If the following equation is properly balanced, what are the coefficients in front of each com- pound? Fe 2 O 4 + SO 2 FeS + O 2 1. 2; 1; 1; 4 2. 2; 2; 1; 4 3. 2; 1; 2; 4 4. 1; 2; 2; 4 correct 5. 1; 2; 1; 4 6. 2; 2; 2; 5 7. 2; 2; 1; 5 8. 1; 2; 1; 2 9. 1; 1; 1; 1 Explanation: The balanced equation is Fe 2 O 4 + 2 SO 2 2 FeS + 4 O 2 Start by looking at the Fe 2 O 4 . This implies 2 molecules of FeS are needed, which then im- plies 2 molecules of SO 2 are needed. Finally, 4 molecules of O 2 are needed. 004 10.0 points What is the percent of nitrogen (N) by mass in a 55.0 g sample of Fe(NO 3 ) 3 ? The molar mass of Fe(NO 3 ) 3 is 241.86 g/mol.

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aleman (ka6539) – homework 2 – Cowley – (90225) 2 1. 5.80% 2. 9.55% 3. 17.4% correct 4. 52.2% 5. 5.80 g 6. 9.55 g 7. 52.2 g 8. 17.4 g Explanation: % N = mass N mass Fe(NO 3 ) 3 × 100% First we need to calculate the mass of N in the sample ? g N = 55 . 0 g Fe(NO 3 ) 3 × 1 mol Fe(NO 3 ) 3 241 . 86 g Fe(NO 3 ) 3 × 3 mol N 1 mol Fe(NO 3 ) 3 × 14 . 007 g N 1 mol N = 9 . 56 g N Now we can calculate the percent of N in the sample % N = 9 . 56 g N 55 . 0 g Fe(NO 3 ) 3 × 100% = 17 . 4% N 005 10.0 points 0.723 moles of ZnO 2 would weigh 1. 70.4 g. correct 2. 45.6 g. 3. 0.00742 g. 4. 62.2 g. 5. 135 g. 6. 58.9 g. Explanation: n ZnO 2 = 0.723 mol ? g = 0 . 723 mol ZnO 2 × 97 . 4 g 1 mol ZnO 2 = 70 . 4 g 006 10.0 points A substance has a molar mass of 60 g/mol with 40.0 percent carbon, 6.7 percent hydro- gen, and 53.3 percent oxygen. What is its empirical formula? 1. C 2 H 2 O 3 2. CHO 2 3. CHO 4. C 2 HO 5. C 3 H 2 O 2 6. CH 2 O correct 7. C 2 H 4 O 2 8. C 2 H 3 O 2 9. C 3 H 3 O 2 Explanation: MW = 60 % C = 40.0% % H = 6.7% % O = 53.3% Assume we have 100 g of this compound. (In fact, we can assume any mass, but choos- ing 100 g makes the math easier.) If we have 100 g of this compound, 40.0 g (40.0%) of it is carbon, 6.7 g (6.7%) of it is hydrogen, and 53.3 (53.3%) of it is oxygen.
aleman (ka6539) – homework 2 – Cowley – (90225) 3 This means that, in this 100 g sample: ? mol of C = 40 . 0 g C × 1 mol C 12 . 011 g C = 3 . 33 mol C ? mol of H = 6 . 7 g H × 1 mol H 1 . 0079 g H = 6 . 6 mol H ? mol of O = 53 . 3 g O × 1 mol O 15 . 9994 g O = 3 . 33 mol O The smallest number of moles here is then 3.33, and we can use this number to detem- ine the empirical (simplest) formula for this compound. For each element, we should now

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